Lesson 5 of 12
42% complete
1. Further Differentiation P2
Further Standard Derivatives
d/dx (tan x) = sec²x d/dx (sec x) = sec x tan x
d/dx (cosec x) = −cosec x cot x d/dx (cot x) = −cosec²x
d/dx (arcsin x) = 1/√(1−x²)
d/dx (arccos x) = −1/√(1−x²)
d/dx (arctan x) = 1/(1+x²)
d/dx [arcsin(x/a)] = 1/√(a²−x²) d/dx [arctan(x/a)] = a/(a²+x²)
📐 Worked Example 1 — Differentiating Reciprocal and Inverse Trig
Differentiate: (a) y = sec(3x) (b) y = arctan(2x) (c) y = x arcsin x (d) y = ln(sec x + tan x)
1
(a) Chain: d/dx[sec u]·3 = sec(3x)tan(3x)·3 = 3 sec(3x) tan(3x)
2
(b) d/dx[arctan(2x)] = 1/(1+(2x)²)·2 = 2/(1+4x²)
3
(c) Product: arcsin x + x·1/√(1−x²) = arcsin x + x/√(1−x²)
4
(d) Let u = sec x + tan x. du/dx = sec x tan x + sec²x = sec x(tan x + sec x) = sec x · u
dy/dx = (1/u)·sec x · u = sec x
(This is the standard result: d/dx[ln|sec x + tan x|] = sec x)
dy/dx = (1/u)·sec x · u = sec x
(This is the standard result: d/dx[ln|sec x + tan x|] = sec x)
Logarithmic Differentiation
Logarithmic differentiation is used when y is a product of many factors raised to powers, or when the exponent contains x. Take ln of both sides, differentiate implicitly, then multiply by y.
📐 Worked Example 2 — Logarithmic Differentiation
Find dy/dx for: (a) y = xˣ (b) y = x²(x+1)³/(2x−1)⁴
1
(a) ln y = x ln x. Differentiate implicitly:
(1/y)(dy/dx) = ln x + x·(1/x) = ln x + 1
dy/dx = xˣ(1 + ln x)
(1/y)(dy/dx) = ln x + x·(1/x) = ln x + 1
dy/dx = xˣ(1 + ln x)
2
(b) ln y = 2ln x + 3ln(x+1) − 4ln(2x−1)
(1/y)dy/dx = 2/x + 3/(x+1) − 8/(2x−1)
dy/dx = y[2/x + 3/(x+1) − 8/(2x−1)] where y = x²(x+1)³/(2x−1)⁴
(1/y)dy/dx = 2/x + 3/(x+1) − 8/(2x−1)
dy/dx = y[2/x + 3/(x+1) − 8/(2x−1)] where y = x²(x+1)³/(2x−1)⁴
2. Integration by Parts P2
Integration by Parts Formula
∫u(dv/dx) dx = uv − ∫v(du/dx) dx
Or: ∫u dv = uv − ∫v du
LIATE rule for choosing u (differentiate u, integrate dv/dx):
Logarithm > Inverse trig > Algebraic > Trig > Exponential
Choose u as the type that appears earlier in LIATE.
📐 Worked Example 3 — Integration by Parts
Find: (a) ∫x eˣ dx (b) ∫x² sin x dx (c) ∫ln x dx (d) ∫arctan x dx
1
(a) u=x, dv=eˣdx → du=dx, v=eˣ
∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + c = eˣ(x−1) + c
∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + c = eˣ(x−1) + c
2
(b) Apply twice. u=x², dv=sin x dx → du=2x dx, v=−cos x
∫x²sin x dx = −x²cos x + ∫2x cos x dx
Now ∫2x cos x dx: u=2x, dv=cos x → du=2dx, v=sin x
= 2x sin x − ∫2 sin x dx = 2x sin x + 2cos x
= −x²cos x + 2x sin x + 2cos x + c
∫x²sin x dx = −x²cos x + ∫2x cos x dx
Now ∫2x cos x dx: u=2x, dv=cos x → du=2dx, v=sin x
= 2x sin x − ∫2 sin x dx = 2x sin x + 2cos x
= −x²cos x + 2x sin x + 2cos x + c
3
(c) ∫ln x dx: u=ln x, dv=dx → du=1/x dx, v=x
= x ln x − ∫x·(1/x) dx = x ln x − ∫1 dx = x ln x − x + c
= x ln x − ∫x·(1/x) dx = x ln x − ∫1 dx = x ln x − x + c
4
(d) ∫arctan x dx: u=arctan x, dv=dx → du=1/(1+x²) dx, v=x
= x arctan x − ∫x/(1+x²) dx = x arctan x − ½ln(1+x²) + c
= x arctan x − ½ln(1+x²) + c
= x arctan x − ∫x/(1+x²) dx = x arctan x − ½ln(1+x²) + c
= x arctan x − ½ln(1+x²) + c
📐 Worked Example 4 — Cyclic Integration by Parts
Find ∫eˣ sin x dx.
1
u=sin x, dv=eˣdx → du=cos x dx, v=eˣ
I = eˣ sin x − ∫eˣ cos x dx
I = eˣ sin x − ∫eˣ cos x dx
2
For ∫eˣ cos x dx: u=cos x, dv=eˣdx → du=−sin x dx, v=eˣ
∫eˣ cos x dx = eˣ cos x − ∫eˣ(−sin x)dx = eˣ cos x + ∫eˣ sin x dx = eˣ cos x + I
∫eˣ cos x dx = eˣ cos x − ∫eˣ(−sin x)dx = eˣ cos x + ∫eˣ sin x dx = eˣ cos x + I
3
Substitute back: I = eˣ sin x − (eˣ cos x + I)
2I = eˣ sin x − eˣ cos x = eˣ(sin x − cos x)
I = ½eˣ(sin x − cos x) + c
2I = eˣ sin x − eˣ cos x = eˣ(sin x − cos x)
I = ½eˣ(sin x − cos x) + c
3. Further Integration Techniques P2
Integrating Partial Fractions
📐 Worked Example 5 — Integration Using Partial Fractions
Find ∫(4x+2)/[(x+3)(x−1)] dx.
1
Partial fractions: (4x+2)/[(x+3)(x−1)] = A/(x+3) + B/(x−1)
4x+2 = A(x−1)+B(x+3): x=1→B=6/4=3/2; x=−3→A=−10/(−4)=5/2
4x+2 = A(x−1)+B(x+3): x=1→B=6/4=3/2; x=−3→A=−10/(−4)=5/2
2
∫[(5/2)/(x+3) + (3/2)/(x−1)] dx
= 5/2 ln|x+3| + 3/2 ln|x−1| + c
= 5/2 ln|x+3| + 3/2 ln|x−1| + c
Integrating Trig Functions Using Identities
Key Integration Results Using Identities
∫sin²x dx = x/2 − sin2x/4 + c (use cos2x = 1−2sin²x)
∫cos²x dx = x/2 + sin2x/4 + c (use cos2x = 2cos²x−1)
∫tan²x dx = tan x − x + c (use tan²x = sec²x−1)
∫sin x cos x dx = sin²x/2 + c = −cos2x/4 + c
∫sec x dx = ln|sec x + tan x| + c
∫cosec x dx = −ln|cosec x + cot x| + c = ln|cosec x − cot x| + c
📐 Worked Example 6 — Trig Integration
Find: (a) ∫sin⁴x dx (b) ∫sin³x dx
1
(a) sin⁴x = (sin²x)² = [½(1−cos2x)]² = ¼(1−2cos2x+cos²2x)
cos²2x = ½(1+cos4x)
sin⁴x = ¼(1−2cos2x+½+½cos4x) = ¼(3/2−2cos2x+½cos4x)
∫sin⁴x dx = ¼[3x/2 − sin2x + sin4x/8] + c
= 3x/8 − sin2x/4 + sin4x/32 + c
cos²2x = ½(1+cos4x)
sin⁴x = ¼(1−2cos2x+½+½cos4x) = ¼(3/2−2cos2x+½cos4x)
∫sin⁴x dx = ¼[3x/2 − sin2x + sin4x/8] + c
= 3x/8 − sin2x/4 + sin4x/32 + c
2
(b) sin³x = sin x(1−cos²x) = sin x − sin x cos²x
∫sin x dx = −cos x; ∫sin x cos²x dx: let u=cos x, du=−sin x dx → ∫−u² du = u³/3
= −cos x + cos³x/3 + c
∫sin x dx = −cos x; ∫sin x cos²x dx: let u=cos x, du=−sin x dx → ∫−u² du = u³/3
= −cos x + cos³x/3 + c
Integration Using Standard Forms Involving arcsin and arctan
Inverse Trig Integration Results
∫1/√(a²−x²) dx = arcsin(x/a) + c (|x| < a)
∫1/(a²+x²) dx = (1/a)arctan(x/a) + c
∫1/√(a²−(bx)²) dx = (1/b)arcsin(bx/a) + c
∫1/((bx)²+a²) dx = (1/(ab))arctan(bx/a) + c
📐 Worked Example 7 — arcsin and arctan Integration
Find: (a) ∫1/√(9−x²) dx (b) ∫1/(4+9x²) dx (c) ∫₀² 1/(x²+4) dx
1
(a) a=3: arcsin(x/3) + c
2
(b) ∫1/(4+9x²) dx = ∫1/[4(1+(9x²/4))] dx = (1/4)∫1/[1+(3x/2)²] dx
= (1/4)·(2/3)arctan(3x/2) + c = (1/6)arctan(3x/2) + c
= (1/4)·(2/3)arctan(3x/2) + c = (1/6)arctan(3x/2) + c
3
(c) ∫₀²1/(x²+4) dx = [(1/2)arctan(x/2)]₀²
= (1/2)arctan(1) − (1/2)arctan(0) = (1/2)(π/4) − 0 = π/8
= (1/2)arctan(1) − (1/2)arctan(0) = (1/2)(π/4) − 0 = π/8
Completing the Square in Integration
📐 Worked Example 8 — Completing the Square
Find: (a) ∫1/(x²+4x+13) dx (b) ∫1/√(8+2x−x²) dx
1
(a) x²+4x+13 = (x+2)²+9
∫1/[(x+2)²+9] dx = (1/3)arctan((x+2)/3) + c
= (1/3)arctan((x+2)/3) + c
∫1/[(x+2)²+9] dx = (1/3)arctan((x+2)/3) + c
= (1/3)arctan((x+2)/3) + c
2
(b) 8+2x−x² = −(x²−2x−8) = −((x−1)²−9) = 9−(x−1)²
∫1/√(9−(x−1)²) dx = arcsin((x−1)/3) + c
= arcsin((x−1)/3) + c
∫1/√(9−(x−1)²) dx = arcsin((x−1)/3) + c
= arcsin((x−1)/3) + c
4. Further Integration by Substitution P2
📐 Worked Example 9 — Trigonometric Substitution
Use the substitution x = 3sinθ to find ∫x²/√(9−x²) dx.
1
x=3sinθ → dx=3cosθ dθ, √(9−x²)=√(9−9sin²θ)=3cosθ
2
∫(9sin²θ)(3cosθ dθ)/(3cosθ) = 9∫sin²θ dθ = 9∫½(1−cos2θ) dθ
= 9[θ/2 − sin2θ/4] + c = 9[θ/2 − sinθcosθ/2] + c
= 9[θ/2 − sin2θ/4] + c = 9[θ/2 − sinθcosθ/2] + c
3
Back-substitute: θ=arcsin(x/3), sinθ=x/3, cosθ=√(9−x²)/3
= 9[arcsin(x/3)/2 − x√(9−x²)/18] + c = (9/2)arcsin(x/3) − x√(9−x²)/2 + c
= 9[arcsin(x/3)/2 − x√(9−x²)/18] + c = (9/2)arcsin(x/3) − x√(9−x²)/2 + c
📐 Worked Example 10 — Definite Integral with Substitution
Use the substitution u = eˣ+1 to find ∫₀^(ln3) eˣ/(eˣ+1)² dx.
1
u=eˣ+1 → du=eˣdx. Limits: x=0→u=2; x=ln3→u=4
2
∫₂⁴ 1/u² du = [−1/u]₂⁴ = −1/4+1/2 = 1/4
5. Improper Integrals P2
Improper integral: An integral where either the limits are infinite, or the integrand is undefined (infinite) at a point within or at the boundary of the interval. To evaluate, replace the infinite limit or problematic point with a parameter t, integrate, then take the limit as t → ∞ (or t → the problematic value).
Types of Improper Integrals
Infinite limit: ∫_a^∞ f(x) dx = lim[t→∞] ∫_a^t f(x) dx
Unbounded integrand: ∫_a^b f(x) dx = lim[t→a⁺] ∫_t^b f(x) dx (if f→∞ at x=a)
Converges if the limit exists and is finite; Diverges otherwise
📐 Worked Example 11 — Improper Integrals
Evaluate: (a) ∫₁^∞ 1/x² dx (b) ∫₁^∞ 1/x dx (c) ∫₀¹ 1/√x dx
1
(a) ∫₁^t x⁻² dx = [−x⁻¹]₁^t = −1/t+1. As t→∞: → 1.
∫₁^∞ 1/x² dx = 1 (converges)
∫₁^∞ 1/x² dx = 1 (converges)
2
(b) ∫₁^t 1/x dx = [ln x]₁^t = ln t. As t→∞: ln t → ∞.
∫₁^∞ 1/x dx diverges
∫₁^∞ 1/x dx diverges
3
(c) Integrand → ∞ at x=0. ∫_ε^1 x^(−½) dx = [2√x]_ε^1 = 2−2√ε. As ε→0⁺: → 2.
∫₀¹ 1/√x dx = 2 (converges)
∫₀¹ 1/√x dx = 2 (converges)
Convergence Quick Guide:
∫₁^∞ 1/xᵖ dx converges if p > 1 (value = 1/(p−1)); diverges if p ≤ 1.
∫₀¹ 1/xᵖ dx converges if p < 1 (value = 1/(1−p)); diverges if p ≥ 1.
Note the boundary p=1 diverges in both cases (gives ln).
∫₁^∞ 1/xᵖ dx converges if p > 1 (value = 1/(p−1)); diverges if p ≤ 1.
∫₀¹ 1/xᵖ dx converges if p < 1 (value = 1/(1−p)); diverges if p ≥ 1.
Note the boundary p=1 diverges in both cases (gives ln).
6. Introduction to Differential Equations P2
Differential equation: An equation relating a function to its derivatives. A first-order ODE involves dy/dx; a second-order involves d²y/dx². P2 covers first-order separable ODEs; P3 covers more types.
Separable Differential Equations
Method — Variables Separable:
1. Rearrange so all y terms (including dy) are on the left, all x terms (including dx) on the right.
2. Integrate both sides.
3. Apply the initial condition (if given) to find the constant of integration.
4. Express y explicitly if possible.
1. Rearrange so all y terms (including dy) are on the left, all x terms (including dx) on the right.
2. Integrate both sides.
3. Apply the initial condition (if given) to find the constant of integration.
4. Express y explicitly if possible.
📐 Worked Example 12 — Separable ODE
Solve dy/dx = xy, given y = 2 when x = 0.
1
Separate: dy/y = x dx
Integrate: ln|y| = x²/2 + c
Integrate: ln|y| = x²/2 + c
2
|y| = e^(x²/2+c) = Ae^(x²/2) where A = eᶜ > 0
y = Ae^(x²/2) (taking A as any non-zero constant)
y = Ae^(x²/2) (taking A as any non-zero constant)
3
y=2 at x=0: 2 = Ae⁰ = A
y = 2e^(x²/2)
y = 2e^(x²/2)
📐 Worked Example 13 — ODE with Partial Fractions
Solve dy/dx = (1−y)/x, given y = ½ when x = 1. Express y in terms of x.
1
Separate: dy/(1−y) = dx/x
Integrate: −ln|1−y| = ln|x| + c → ln|1−y| = −ln x + c (x > 0, so |x|=x)
Integrate: −ln|1−y| = ln|x| + c → ln|1−y| = −ln x + c (x > 0, so |x|=x)
2
|1−y| = e^(c−ln x) = A/x where A = eᶜ > 0
1−y = A/x → y = 1 − A/x
1−y = A/x → y = 1 − A/x
3
y=½ at x=1: ½ = 1−A → A = ½
y = 1 − 1/(2x)
y = 1 − 1/(2x)
📐 Worked Example 14 — Modelling with a Differential Equation
A population P grows at a rate proportional to its current size. Initially P = 500. After 10 years P = 1200. Find P as a function of t, and find t when P = 5000.
1
dP/dt = kP → dP/P = k dt → ln P = kt + c → P = Aeᵏᵗ
2
P(0)=500: A=500. P(10)=1200: 1200=500e^(10k)
e^(10k)=2.4 → k = ln(2.4)/10 ≈ 0.0875
P = 500e^(0.0875t)
e^(10k)=2.4 → k = ln(2.4)/10 ≈ 0.0875
P = 500e^(0.0875t)
3
P=5000: 5000=500e^(0.0875t) → e^(0.0875t)=10 → t=ln(10)/0.0875
t ≈ 26.3 years
t ≈ 26.3 years
📝 Exam Practice Questions
Q1 [4 marks] — Find: (a) ∫x ln x dx (b) ∫x²eˣ dx
(a) u=ln x, dv=x dx → du=1/x dx, v=x²/2
∫x ln x dx = (x²/2)ln x − ∫(x²/2)(1/x)dx = (x²/2)ln x − ∫x/2 dx
= (x²/2)ln x − x²/4 + c = x²(2ln x − 1)/4 + c
(b) Apply twice: ∫x²eˣdx = x²eˣ − 2∫xeˣdx = x²eˣ − 2(xeˣ−eˣ) + c
= eˣ(x²−2x+2) + c
∫x ln x dx = (x²/2)ln x − ∫(x²/2)(1/x)dx = (x²/2)ln x − ∫x/2 dx
= (x²/2)ln x − x²/4 + c = x²(2ln x − 1)/4 + c
(b) Apply twice: ∫x²eˣdx = x²eˣ − 2∫xeˣdx = x²eˣ − 2(xeˣ−eˣ) + c
= eˣ(x²−2x+2) + c
Q2 [4 marks] — Find ∫(3x+1)/[(x−1)(x²+1)] dx.
Partial fractions: (3x+1)/[(x−1)(x²+1)] = A/(x−1) + (Bx+C)/(x²+1)
3x+1=A(x²+1)+(Bx+C)(x−1). x=1: 4=2A→A=2.
Expand: 3x+1=2x²+2+Bx²−Bx+Cx−C. Coeff x²: 0=2+B→B=−2. Coeff x: 3=−B+C→C=1. Const: 1=2−C→C=1 ✓
∫[2/(x−1) + (−2x+1)/(x²+1)]dx = 2ln|x−1| − ln(x²+1) + arctan x + c
= 2ln|x−1| − ln(x²+1) + arctan x + c
3x+1=A(x²+1)+(Bx+C)(x−1). x=1: 4=2A→A=2.
Expand: 3x+1=2x²+2+Bx²−Bx+Cx−C. Coeff x²: 0=2+B→B=−2. Coeff x: 3=−B+C→C=1. Const: 1=2−C→C=1 ✓
∫[2/(x−1) + (−2x+1)/(x²+1)]dx = 2ln|x−1| − ln(x²+1) + arctan x + c
= 2ln|x−1| − ln(x²+1) + arctan x + c
Q3 [4 marks] — Evaluate ∫₀^(π/2) sin²x cos³x dx.
Write cos³x = cos x(1−sin²x):
∫₀^(π/2) sin²x cos x(1−sin²x) dx = ∫₀^(π/2)(sin²x cos x − sin⁴x cos x) dx
Let u=sin x, du=cos x dx. Limits: 0→0, π/2→1
∫₀¹(u²−u⁴)du = [u³/3−u⁵/5]₀¹ = 1/3−1/5 = 2/15
∫₀^(π/2) sin²x cos x(1−sin²x) dx = ∫₀^(π/2)(sin²x cos x − sin⁴x cos x) dx
Let u=sin x, du=cos x dx. Limits: 0→0, π/2→1
∫₀¹(u²−u⁴)du = [u³/3−u⁵/5]₀¹ = 1/3−1/5 = 2/15
Q4 [4 marks] — Solve the differential equation dy/dx = e^(x−y), given y = 0 when x = 0. Find y when x = 1.
dy/dx = eˣ·e^(−y). Separate: eʸ dy = eˣ dx
Integrate: eʸ = eˣ + c
y=0, x=0: 1=1+c → c=0
eʸ = eˣ → y = x
When x=1: y = 1
Integrate: eʸ = eˣ + c
y=0, x=0: 1=1+c → c=0
eʸ = eˣ → y = x
When x=1: y = 1
Q5 [4 marks] — Determine whether ∫₁^∞ xe^(−x²) dx converges. If so, find its value.
∫₁^t xe^(−x²) dx. Let u=x², du=2x dx → x dx=du/2
= ∫₁^(t²) ½e^(−u) du = [−½e^(−u)]₁^(t²) = −½e^(−t²) + ½e^(−1)
As t→∞: e^(−t²)→0
∫₁^∞ xe^(−x²) dx = ½e^(−1) = 1/(2e) — converges.
= ∫₁^(t²) ½e^(−u) du = [−½e^(−u)]₁^(t²) = −½e^(−t²) + ½e^(−1)
As t→∞: e^(−t²)→0
∫₁^∞ xe^(−x²) dx = ½e^(−1) = 1/(2e) — converges.
Q6 [5 marks] — A curve has the property that dy/dx = y(3−y)/6. Given y = 1 when x = 0: (a) Solve the ODE to find y in terms of x. (b) Find the value of y as x→∞.
(a) Separate: 6dy/[y(3−y)] = dx
Partial fractions: 6/[y(3−y)] = 2/y + 2/(3−y) (since 6/(3y−y²) = A/y+B/(3−y): A=2, B=2)
∫(2/y + 2/(3−y))dy = ∫dx → 2ln y − 2ln(3−y) = x + c → 2ln[y/(3−y)] = x+c
y=1, x=0: 2ln(1/2)=c → c=−2ln2
2ln[y/(3−y)] = x−2ln2 → ln[y/(3−y)] = x/2−ln2 → y/(3−y) = e^(x/2)/2
2y = (3−y)e^(x/2) → y(2+e^(x/2)) = 3e^(x/2)
y = 3e^(x/2)/(2+e^(x/2)) = 3/(2e^(−x/2)+1)
(b) As x→∞: e^(−x/2)→0 → y→3
Partial fractions: 6/[y(3−y)] = 2/y + 2/(3−y) (since 6/(3y−y²) = A/y+B/(3−y): A=2, B=2)
∫(2/y + 2/(3−y))dy = ∫dx → 2ln y − 2ln(3−y) = x + c → 2ln[y/(3−y)] = x+c
y=1, x=0: 2ln(1/2)=c → c=−2ln2
2ln[y/(3−y)] = x−2ln2 → ln[y/(3−y)] = x/2−ln2 → y/(3−y) = e^(x/2)/2
2y = (3−y)e^(x/2) → y(2+e^(x/2)) = 3e^(x/2)
y = 3e^(x/2)/(2+e^(x/2)) = 3/(2e^(−x/2)+1)
(b) As x→∞: e^(−x/2)→0 → y→3
Exam Tip: Logistic-type ODEs dy/dx=ky(L−y) always have the long-term behaviour y→L as x→∞ (when y starts between 0 and L). Recognising this quickly can check your answer without fully solving.